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sm4 的s盒_SM4国密算法Java版

發布時間:2025/3/20 java 47 豆豆
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根據 國密SM4 文檔 編寫的一個Java 加密解密樣例package javasm4;

/**

*

* @author Jeen

*/

public class JavaSM4 {

public static int[] key = new int[4];//密鑰

public static int[] temp = new int[4];//中間量 存儲運算結果

public static int[] rkey = new int[32];//輪密鑰

public static int[] fk = {0xa3b1bac6, 0x56AA3350, 0x677d9197, 0xb27022dc}; //系統參數

public static int[] ck = {//ck[i][j] = (4i+j)×7(mod 256) i=0->31 j=0->3 //固定參數

0x00070e15, 0x1c232a31, 0x383f464d, 0x545b6269,

0x70777e85, 0x8c939aa1, 0xa8afb6bd, 0xc4cbd2d9,

0xe0e7eef5, 0xfc030a11, 0x181f262d, 0x343b4249,

0x50575e65, 0x6c737a81, 0x888f969d, 0xa4abb2b9,

0xc0c7ced5, 0xdce3eaf1, 0xf8ff060d, 0x141b2229,

0x30373e45, 0x4c535a61, 0x686f767d, 0x848b9299,

0xa0a7aeb5, 0xbcc3cad1, 0xd8dfe6ed, 0xf4fb0209,

0x10171e25, 0x2c333a41, 0x484f565d, 0x646b7279};

private static int[] sbi = { //sbi 用于8位置換的數組 sbox

0xd6,0x90,0xe9,0xfe,0xcc,0xe1,0x3d,0xb7,0x16,0xb6,0x14,0xc2,0x28,0xfb,0x2c,0x05,

0x2b,0x67,0x9a,0x76,0x2a,0xbe,0x04,0xc3,0xaa,0x44,0x13,0x26,0x49,0x86,0x06,0x99,

0x9c,0x42,0x50,0xf4,0x91,0xef,0x98,0x7a,0x33,0x54,0x0b,0x43,0xed,0xcf,0xac,0x62,

0xe4,0xb3,0x1c,0xa9,0xc9,0x08,0xe8,0x95,0x80,0xdf,0x94,0xfa,0x75,0x8f,0x3f,0xa6,

0x47,0x07,0xa7,0xfc,0xf3,0x73,0x17,0xba,0x83,0x59,0x3c,0x19,0xe6,0x85,0x4f,0xa8,

0x68,0x6b,0x81,0xb2,0x71,0x64,0xda,0x8b,0xf8,0xeb,0x0f,0x4b,0x70,0x56,0x9d,0x35,

0x1e,0x24,0x0e,0x5e,0x63,0x58,0xd1,0xa2,0x25,0x22,0x7c,0x3b,0x01,0x21,0x78,0x87,

0xd4,0x00,0x46,0x57,0x9f,0xd3,0x27,0x52,0x4c,0x36,0x02,0xe7,0xa0,0xc4,0xc8,0x9e,

0xea,0xbf,0x8a,0xd2,0x40,0xc7,0x38,0xb5,0xa3,0xf7,0xf2,0xce,0xf9,0x61,0x15,0xa1,

0xe0,0xae,0x5d,0xa4,0x9b,0x34,0x1a,0x55,0xad,0x93,0x32,0x30,0xf5,0x8c,0xb1,0xe3,

0x1d,0xf6,0xe2,0x2e,0x82,0x66,0xca,0x60,0xc0,0x29,0x23,0xab,0x0d,0x53,0x4e,0x6f,

0xd5,0xdb,0x37,0x45,0xde,0xfd,0x8e,0x2f,0x03,0xff,0x6a,0x72,0x6d,0x6c,0x5b,0x51,

0x8d,0x1b,0xaf,0x92,0xbb,0xdd,0xbc,0x7f,0x11,0xd9,0x5c,0x41,0x1f,0x10,0x5a,0xd8,

0x0a,0xc1,0x31,0x88,0xa5,0xcd,0x7b,0xbd,0x2d,0x74,0xd0,0x12,0xb8,0xe5,0xb4,0xb0,

0x89,0x69,0x97,0x4a,0x0c,0x96,0x77,0x7e,0x65,0xb9,0xf1,0x09,0xc5,0x6e,0xc6,0x84,

0x18,0xf0,0x7d,0xec,0x3a,0xdc,0x4d,0x20,0x79,0xee,0x5f,0x3e,0xd7,0xcb,0x39,0x48};

public static void main(String[] args)

{

JavaSM4 sm = new JavaSM4();

int[] msg = {0x01234567, 0x89abcdef, 0xfedcba98, 0x76543210};

int[] smsg = {0x595298c7, 0xc6fd271f, 0x0402f804, 0xc33d3f66};

key[0] = 0x01234567;

key[1] = 0x89abcdef;

key[2] = 0xfedcba98;

key[3] = 0x76543210;

int j=0,n=1000000;

long startTime = System.currentTimeMillis(); //獲取開始時間

for(j=0; j

{

//msg = sm4(msg,1);//加密運算

smsg = sm4(smsg,0);//解密運算

}

long endTime = System.currentTimeMillis(); //獲取結束時間

System.out.println("執行SM4加密運算"+n+"次時間: "+(endTime-startTime)+"ms");

}

private static int[] sm4(int[] t,int s) //s!=0 時為加密運算,s=0時為解密運算

{

rkey = initrk();

if(s == 0)

{

rkey = r(rkey);

}

int[] x = new int[36];

x[0] = t[0];

x[1] = t[1];

x[2] = t[2];

x[3] = t[3];

int i;

for(i=0;i<32;i++)

{

x[i+4] = f(x[i],x[i+1],x[i+2],x[i+3],rkey[i]);

}

x = r(x);

temp[0] = x[0];

temp[1] = x[1];

temp[2] = x[2];

temp[3] = x[3];

return temp;

}

private static int[] initrk()

{

int i;

int[] k = new int[36];

int[] rk = new int[32];

k[0] = key[0] ^ fk[0];

k[1] = key[1] ^ fk[1];

k[2] = key[2] ^ fk[2];

k[3] = key[3] ^ fk[3];

for(i=0;i<32;i++)

{

rk[i] = k[i+4] = k[i] ^ tn(k[i+1]^k[i+2]^k[i+3]^ck[i]);

}

return rk;

}

private static int[] r(int[] x)

{

int[] t = new int[x.length];

int i;

for(i=0; i

{

t[i] = x[x.length - 1 -i];

}

return t;

}

private static int f(int x0,int x1,int x2,int x3,int k)

{

return (x0 ^ t(x1 ^ x2 ^ x3 ^ k));

}

private static int t(int ta)

{

return l(tj(ta));

}

private static int tn(int ta)

{

return ln(tj(ta));

}

private static int l(int temp)

{

return temp ^ Px(temp,2) ^ Px(temp,10) ^ Px(temp,18) ^ Px(temp,24);

}

private static int ln(int temp)

{

return temp ^ Px(temp,13) ^ Px(temp,23);

}

private static int tj(int a)

{

byte[] b = new byte[4];

byte[] c = new byte[4];

c = intToBytes(a);

b[0] = sbox(c[0]);

b[1] = sbox(c[1]);

b[2] = sbox(c[2]);

b[3] = sbox(c[3]);

a = bytesToInt(b[0],b[1],b[2],b[3]);

return a;

}

private static byte sbox(byte a) //S盒 8 bit 置換

{

int t = (a << 24) >>> 24;

return (byte)sbi[t];

}

private static int Px(int x,int n) //整型循環左移運算 n <=32 //java中沒有無符號整型,需要注意移位后的填充符

{

return ((x<>>(32-n)));

}

private static int bytesToInt(byte b0,byte b1,byte b2,byte b3) // int = 4 * byte = 32 bit unsigned

{

int tint = 0;

int temp = b0 << 24;

tint = temp;

temp = (b1 << 24) >>> 8;

tint |= temp;

temp = (b2 << 24) >>> 16;

tint |= temp;

temp = (b3 << 24) >>> 24;

tint |= temp;

return tint;

}

private static byte[] intToBytes(int i)

{

byte[] tbyte = new byte[4];

tbyte[0] = (byte)(i >>> 24);

tbyte[1] = (byte)((i<<8)>>>24);

tbyte[2] = (byte)((i<<16)>>>24);

tbyte[3] = (byte)((i<<24)>>>24);

return tbyte;

}

}

TODO: 1文件信息流處理 ? 2長度不足的填充算法

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